Two bodies of masses $m_1$ and $m_2$ initially at rest at infinite distance apart move towards each other under gravitational force of attraction. Their relative velocity of approach when they are separated by a distance $r$ is ( $\mathrm{G}=$ universal gravitational constant.)

Question

Two bodies of masses $m_1$ and $m_2$ initially at rest at infinite distance apart move towards each other under gravitational force of attraction. Their relative velocity of approach when they are separated by a distance $r$ is ( $\mathrm{G}=$ universal gravitational constant.), $\left[\frac{2 G\left(m_1-m_2\right)}{r}\right]^{1 / 2}$, $\left[\frac{2 G\left(m_1+m_2\right)}{r}\right]^{1 / 2}$, $\left[\frac{r}{2 G\left(m_1 m_2\right)}\right]^{1 / 2}$, $\left[\frac{r}{2 G} m_1 m_2\right]^{1 / 2}$

MARKS App · Accepted Answer

Initially when the two masses are at an infinite distance from each other, their gravitational potential energy is zero. When they are at a distance $r$ from each, the gravitational $P E$ is $\mathrm{PE}=\frac{-G m_1 m_2}{r^2}$ The minus sign-indicates that there is a decrease in PE. This gives rise to an increase in kinetic energy. If $v_1$ and $v_2$ are their respective velocities when they are a distance $r$ apart then, from the law of conservation of energy, we have $\frac{1}{2} m_1 v_1^2=\frac{G m_1 m_2}{r} \text { or } v_1=\sqrt{\frac{2 G m_2}{r}}$ and $\frac{1}{2} m_2 v_2^2=\frac{G m_1 m_2}{r}$ or $v_2=\sqrt{\frac{2 G m_1}{r}}$ Therefore, their relative velocity of approach is $\begin{aligned} v_1+v_2 & =\sqrt{\frac{2 G m_2}{r}}+\sqrt{\frac{2 G m_1}{r}}=\sqrt{\frac{2 G}{r}\left(m_1+m_2\right)} \ & =\left(\frac{2 G\left(m_1 \times m_2\right)}{r}\right)^{\frac{1}{2}} \end{aligned}$

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